All categories

4 years ago

How many moles of Pb(NO3)2 are required

Chemistry

1 answer:

yarga [219]4 years ago

6 0

Answer:

12 moles Pb(NO₄)₂ needed.

Explanation:

3Pb(NO₃)₂ +2AlCl₃ => 3PbCl₂ + 2Al(NO₃)₃

Given => ? moles 8 moles

from reaction stoichiometry, 2 moles AlCl₃ requires 3 moles Pb(NO₄)₂ then 8 moles AlCl₃ requires 3/2(8) moles of the Pb(NO₄)₂ => 12 moles Pb(NO₄)₂ needed.

You might be interested in

Which chemical equation correctly represents the reaction that takes place when nitrogen gas and hydrogen gas are formed as ammo

Dima020 [189]

Answer:

D. 2NH₃(g) ⟶ N₂(g) + 3H₂(g)

Explanation:

A chemical equation must show the correct formulas for reactants and products.

A and B are wrong, because they have NH₃ as a product.

C is wrong. H₃ does not exist.

A is correct. The equation for the decomposition of ammonia is

2NH₃(g) ⟶ N₂(g) + 3H₂(g)

5 0

4 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

6 0

3 years ago

Isotopes have

Hatshy [7]

A. isotopes only hve different mass which is caused by additional/lesser neutron. protons and electrons number are fixed

5 0

3 years ago

Please only answer if you know the actual thing, don't search it up! Are these correct? :P

makkiz [27]

Hey buddy I am here to help!

1. C

2. A

3. A & B

4. C

5. C

6. A

7. A

8. A & C

Hope it helps!

Plz mark brainlist!

4 0

3 years ago

Read 2 more answers

28.

OlgaM077 [116]

Answer:

i don't know

Explanation:

4 0

2 years ago